![]() It is not now possible by this means to change q to unity, but it is possible to ensure that the vector m a (which is the coefficient of largest degree of η in (G2)) becomes perpendicular to the vector n a, i.e. The metric (Gl), (G2) also allows one arbitrary transformation in this case. ![]() The investigation of the cases in which s 3 is not the biggest of the three numbers s 1, s 2, s 3 reduces to the investigation performed above. Thus the metric (G1), (G2) contains only three physically arbitrary functions of the coordinates x, y, ζ. It is possible for example to change the coefficient q in g 33 into unity by this transformation. One should bear in mind here that the allowed transformation must maintain the situation in which the singularity of the metric is at η = 0, and g 33 contains a higher power of η (than in g ab). Besides this the metric (G1), (G2) allows one more transformation (containing one arbitrary function of the coordinates x, y, ζ) without its form being changed. Together with (G13) we have, consequently, four relations between eight functions ( q, two components of each of the vectors l a, m a, n a and one of the numbers s 1, s 2, s 3). Terms ∼ t 2 s 3–1 in these equations give three relations (equal to the number of equations), connecting the functions of the coordinates ‘ x, y, ζ appearing in (G2) one with another. Of these the first identically cancel each other because of the relations (G3) between the numbers s 1, s 2, s 3. Finally the substitution of the metric (G2) together with the condition (G13) leads to the appearance in eqs (G11), (G12) of terms of orders t 2 s 3–1 In t and t 2 s 3–1. Is such a condition (making terms ∼η / 2( s 1 − s 2) in the quantities λ a b equal to zero). How do the feasible region and the solution change? How do the feasible region and the solution change? (B) Modify the script to add a constraint m 2 ≥ 0.2. (A) Modify the MatLab gda07_07 script so that the last constraint is m 1 ≥ 1.2. (B) Run a series of test cases for a range of values of m 1 and comment upon the results. (A) Write a MatLab script that generates synthetic observed data (including Gaussian-distributed noise) for a particular choice of m 1, estimates the model parameters using both simple least squares and nonnegative least squares, plots the observed and predicted data, and outputs the total error. 7.4Ĭonsider fitting a cubic polynomial d i = m 1 + m 2 z i + m 3 z i 2 + m 4 z i 3 to N = 20 data d i obs, where the zs are evenly spaced on the interval (0,1), where m 2 = m 3 = m 4 = 1, and where m 1 is varied from − 1 to 1, as described below. You may wish to refer to MatLab script gda07_03 for the definition of G). Then examine the size of the elements of a. How large a contribution do the null vectors make to the true solution m true? (Hint: Write the model parameters as a linear combination of all the eigenvectors V by writing m true = Va, where a are coefficients, and then solving for a. ![]() (B) Suppose that the elements of m true are drawn from a uniform distribution between 0 and 1. (A) Compute and plot the null vectors for the data kernel shown in Figure 7.4A. In order to insure that W e = D T D has an inverse, which is required by the transformation, you should make D square by adding two rows, one at the top and the other at the bottom, that constrain the first and last model parameters to known values. How different is this solution from the one given in the figure? Compare the two prediction errors. Then, find the minimum-length solution (or the natural solution) m′ est. First, transform the weighted problem Gm = d into an unweighted one G′ m′ = d′ using the transformation given by Equation (7.23). Modify the weighted damped least squares “gap-filling” problem of Figure 3.10 ( MatLab script gda03_09) so that the a priori information is applied only to the part of the solution in the null space. Why? Show that the two sets are equivalent. (A) The null vectors are different from given in Equation (6.5). Use MatLab 's svd() function to compute the null vectors associated with the data kernel G = . This problem builds upon the discussion in Section 6.2.
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